Question

An electron moves to the left with a velocity of 9.6 x 10^5 m/s and has a magnetic force of 2.6 x 10^-15 N exerted on it. What is the magnitude of the magnetic field? Assume the charge has a value of 1.60 x 10^-19 C.

Answer 1

Answer:

Magnetic field, B = 0.016 Tesla

Explanation:

It is given that,

Velocity of electron, $v=9.6\times 10^5\ m/s$

Magnetic force, $F=2.6\times 10^{-15}\ N$

Charge, $q=1.6\times 10^{-19}\ N$  

The magnetic force is given by :

$F=qvB$

$B=\dfrac{F}{qv}$

$B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}$

$B=0.016\ T$

So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.    

Answer 2

Answer:

The magnitude of the magnetic field strength is  16.927 mT

Explanation:

Given;

velocity of the electron, v = 9.6 x 10⁵ m/s

magnitude of the force exerted on the electron, F =  2.6 x 10⁻¹⁵ N

charge of electron, q = 1.60 x 10⁻¹⁹ C

The force exerted on the electron in magnetic field is given by charge times the vector product of velocity and magnetic field strength.

F = q(v x B)

where;

B is the magnetic field strength

$B = \frac{F}{qv}$

Substitute the given values of force, charge and velocity and calculate the magnetic field strength:

$B = \frac{2.6*10^{-15}}{(1.6*10^{-19})(9.6*10^5)}} \\\\B = 1.6927*10^{-2} \ T\\\\B = 16.927 \ mT$

Therefore, the magnitude of the magnetic field strength is 16.927 mT